Draw any line OR, and take M in it so that the measure of OM is unity.

Draw MP perpendicular to OM, and take P so that the measure of MP is a. Join OP. Then the angle ROP is the

Thus an angle ROP has been drawn so that tan ROP=a.

117. DEFINITION. One angle is called the complement of another, when the two angles added together make up a right angle.

Example 1. The complement of A is (90° – A).

=

Example 2. The complement of 190° is (90° – 190o) — — 100o. For 1900+(900-1900)=90°.

118. To prove that the sine of an angle A is equal to the cosine of its complement (90° - A).

[This is true whatever be the magnitude of A, as will be proved later on.]

If A be less than 90', let ROP be A (see last figure).

Draw PM perpendicular to OR.

Then since PMO = 90°, therefore POM +0PM = 90° and therefore OPM (90° - A).

=

Now, sin A =

MP

ОР

= cos OPM = cos (90° — A). Q.E.D.

EXAMPLES. XIX.

(1) Draw an angle whose sine is .
(2) Draw an angle whose cosecant is 2.

(3) Draw an angle whose tangent is 2.

(4) Can an angle be drawn whose tangent is 431 ? (5) Can an angle be drawn whose cosine is ?

(6) Can an angle be drawn whose secant is 5? (7) Find the complements of

ON THE SOLUTION OF TRIGONOMETRICAL EQUATIONS.

119. A TRIGONOMETRICAL equation is an equation in which there is a letter, such as 0, which stands for an angle of unknown magnitude,

The solution of the equation is the process of finding an angle which, if it be substituted for 0, satisfies the equation.

Example 1. Solve cos 0=.

This is a Trigonometrical equation. To solve it we must find some angle such that its cosine is

We know that cos 60o=.

Therefore if 60° be put for the equation is satisfied,

.. 0=60o is a solution of the equation.

Example 2. Solve sin - cosec 0+3=0.

The usual method of solution is to express all the Trigonometrical Ratios in terms of one of them.

This is an equation in which 6, and therefore sin is unknown. It will be convenient if we put x for sin 0, and then solve the equation for x as an ordinary algebraical equation. Thus we get

The value - 2 is inadmissible for sin 6, for there is no angle whose sine is numerically greater than 1 (Art. 115),

Therefore one angle which satisfies this equation for is 30o.

Example 3. To solve the equation cosec – cot2 + 1=0.

therefore 30o is one angle which satisfies the equation for 0.

EXAMPLES. XX.

Find one angle which satisfies each of the following equations,

** MISCELLANEOUS EXAMPLES. XXI.

(1) Prove that 3 sin 600 - 4 sin3 60°-4 cos3 300-3 cos 30o. (2) Prove that tan 30o (1 + cos 30° + cos 60o)=sin 30o + sin 60o. (3) If 2 cos2 6-7 cos 0+3=0, show there is only one value of cos 0.

(4) Find cos from the equation 8 cos2 0 - 8 cos 0+1=0.

(5) Find sin from the equation 8 sin2 0 – 10 sin @+3=0, and prove that one value of ✔ is

(6) Find tan 6 from the equation 12 tan2 0 - 13 tan 0+3=0. (7) If 3 cos2 0 +2.√3. cos 0=5, show that there is only one

value of cose, and that one value of

(8) Prove that the value of sin1

always the same.

is

π

6

+ cos1 + 2. sin2. cos2 0 is

(9) Simplify cos1 A +2. sin2 A. cos2 A.

(10) Express sino A + cos® A in terms of sin2 A and powers of sin2 A.

(11) Express 1+tan1 Ø in terms of cos 0 and its powers.

(13) Express (sec A-tan A)2 in terms of sin A.

(14) Trace the changes in the magnitude of cosec ◊ as @ increases from 0 to

π

2

(15) Trace the changes in the magnitude of cot as e de